Number Representations Proof:
In order to understand this proof, it is important for us to express numbers in decimal
representation. In all our general math classes, we are used to expressing numbers such as 1263.00 in base
10 such that 1263.00=126310. So, how can we prove the “9 trick” that many of us learned at such an early
age?
The “9’s trick” states that if the sum of the digits of a number n is a multiple of 9, then the
original number n is also a multiple of 9; such that if we call the sum of the digits s, then if s mod 9=0,
then n is a multiple of 9.
For example, we will let n=4518 such that s=4+5+1+8. In this sense, we must evaluate
(4+5+1+8) mod 9. When we do so we will find:
(4+5+1+8) mod 9=(18) mod 9=0
So, with this knowledge we can conclude that 4518 is a multiple of 9 and 9|n.
Now, based on our knowledge of numbers and basic arithmetic we must prove why this rule
works. As specified above, we will take some integer n, and let the sum of its digits equal s. In this sense
we can evaluate n as the sum of digits*some power of 10. So, if we let d=the units and k equal the
numerical value of the digit, we will see the following:
For some integer n:
n = dk10k + dk-110k-1 + dk-210k-2 + … + d1101 +d0100 = dk10k + dk-110k-1 + dk-210k-2 + … + d110 +
d0(1)
In order to make this more clear, we will let n=4518. So, if we evaluate 4518 in the instance of d’s
multiplied by powers of 10 we will find:
n = 4518
= 4* + 5* + 1* + 8*
= 4*1000 + 5*100 + 1*10 + 8
= 4000 + 500 + 10 + 8
n= 4518
In this example we have been able to show how an integer is broken up in this way. Now, we can apply
this to our general formulation which states:
n = dk10k + dk-110k-1 + dk-210k-2 + … + d1101 + d0100 = dk10k + dk-110k-1 + dk-210k-2 + … + d110 +
d0(1)
Based upon this, we know that the sum of the digits (s) can be written as the following:
s = dk + dk-1 + dk-2 + … + d1 + d0
Now with our established values for n and s based upon any given integer we can prove that 9 is a divisor
of n if s is a multiple of 9. We can demonstrate this by using the distributive laws and our knowledge of
basic arithmetic by evaluating n-s in the following way:
n = dk10k + dk-110k-1 + dk-210k-2 + … + d110 + d0
s = dk + dk-1 + dk-2 + … + d1 +d0
n-s = (dk10k-dk) + (dk-110k-1-dk-1) + (dk-210k-2-dk-2) + … + (d110-d1) + (d0-d0)
= dk(10k-1) + dk-1(10k-1-1) + dk-2(10k-2-1) + … + d1(10-1)
With this evaluation we can see that each digit, dα, is being multiplied by a (10β-1). In this sense, we can
use our knowledge of the algebraic laws and roots of polynomials to conclude that a factor of (10β-1) is 9.
So, we can see that the entirety of n-s is being multiplied by some multiple of 9, which we will call w. So,
based on the algebraic laws we can see the following:
n-s = 9w
s = n-9w
n = s+9w
So, because of the basis that if either n or s is divisible by 9, the other must be, n and s must both be
divisible by 9!
In conclusion, we have proved why we can tell if any integer is divisible by 9 based on the
summation of its digits.